Rob Appleby, University of Manchester/Cockcro: Ins<tute

Transcription

1 INTRODUCTION TO BEAM DYNAMICS Rob Appleby, University of Manchester/Cockcro: Ins<tute

2 Part One Introduc<on Let s look at a synchrotron Periodicity and stability Hierarchy of beam descripfons SHM and the pendulum Newton and Hamilton Part Two - Hill s equa<ons Beams and magnets DerivaFon of Hill s equafons The transfer matrix approach Matrix properfes Part Three LaFce func<ons The Courant- Synder formalism How to transform the lajce funcfons Stability Tune Part Four Op<cs and lafce design The FODO cell LaJce design Mini beta inserfons Principles of lajce design OpFcal structures

3 Part Five- Errors in our lafces Field errors Closed orbit distorfon Tune ships Part Six Real par<cles Dispersion Momentum compacfon ChromaFcity EmiQance Part Seven Longitudinal dynamics The principle of phase stability The pill box cavity and real cavifes Longitudinal dynamics Buckets and bunches Part Eight Radia<on effects Synchrotron radiafon Damping effects (directed reading)

4 Learning material for this course Course Material Books CI Courses ( hqp:// educafon.htm) Linear Dynamics (Andy Wolski) Accelerator Physics (Rob Appleby) Linear OpFcs and LaJce Design (Bernhard Holzer) OpFcs codes (Hywel Owen) RF (Roger Jones, Graeme Burt) RelaFvity (Kai Hock) ElectromagneFsm And so.. CAS (hqp:// And many more. Lee, Accelerator physics Wiedemann, ParFcle accelerator physics Chao and Tigner, The handbook Wille, The physics of parfcle accelerators Forest, Beam dynamics Wangler, RF linear accelerators And many more. (available from all good booksellers!) (and amazon)

5 The pathway of an accelerator physicist Special relafvity ElectromagneFsm Classical mechanics Beam dynamics Transverse beam dynamics Longitudinal beam dynamics RF design CollecFve effects Magnet design Non- linear dynamics OpFcal design OperaFons + many others

6 PART ONE - INTRODUCTION

7 Largest storage ring: The Solar System astronomical unit: average distance earth-sun 1AE 150 *10 6 km Distance Pluto-Sun 40 AE AE

8 A pendulum and a synchrotron Let s look at a pendulum. x

9 A synchrotron (PS)

10 DIAMOND light source

11 An alien looks Now imagine you are an alien and come to earth to look at the PS What observafons do you make and what would impress you?

12 An alien looks 1) The parfcles spend a long Fme in the ring. The mofon is STABLE 2) The machine is periodic 3) The parfcle mofon does not have the same periodicity as the machine on a parfcle by parfcle basis but the envelope of the mofon follows the machine periodicity. These are observafons we will study and explain using beam dynamics I would add one more, which will become clearer later 4) The mofon is symplecfc in hadron machines and nearly symplecfc in electron machines (but also slightly stochasfc)

13 The LHC

14 LHC parameters LHC design parameters Nominal LHC parameters Beam injection energy (TeV) 0.45 Beam energy (TeV) 7.0 Number of particles per bunch 1.15 x Number of bunches per beam 2808 Max stored beam energy (MJ) 362 Norm transverse emittance (μm rad) 3.75 Colliding beam size (μm) 16 Bunch length at 7 TeV (cm) How do we produce ~3000 proton bunches of 450 GeV? - How do we accelerate them to higher energies?

15 The closed orbit of the LHC There is a closed design orbit around the ring, and all parfcles oscillate around this design orbit by a small amount (~ mm) We need to focus the beam around this closed orbit

16 The life<me of the LHC

17 HERA

18 What is beam dynamics? We do beam dynamics to understand the mofon of parfcles in linear and circular accelerators, to Understand the fundamentals of exisfng machines OpFmise and commission accelerators Design new machines e.g. a new collider Design novel machines e.g. a non- scaling FFAG How do we do this? The fundamental tool of a person doing beam dynamics is knowing how to calculate the mofon of a charged parfcle in a real electromagnefc field Magneto- stafc configurafons Time dependent fields OpFcs What are the approximafons used? How do the parfcles interact with the surroundings? Do the parfcles in a bunch interact with themselves? The most basic quesfon is how do I represent the beam passing through the accelerator in my beam dynamics language? This leads us to a hierarchy of beam descripfons

19 A hierarchy of beam descrip<ons I Think of a box of gas We can think of this system as being described by several numbers The pressure (P) The temperature (T) The volume (V) The number of moles (n) An equafon of state relates these quanffes together An ideal gas obeys the ideal gas law PV=nRT Which relates the state variables to each other and tell us how they change. e.g. an isobaric change of state

20 A hierarchy of beam descrip<ons II Our gas is also made up of a collecfon of gas molecules, each with a posifon and a momentum in every degree of freedom of the system Each molecule has a speed and a kinefc energy (translafonal energy) This is a microscopic view of our gas in a box, and an equally valid view The two pictures are related.. K tr = 3 2 nrt So K tr (the average kinefc energy) is directly proporfonal to the macroscopic temperature of the gas

21 Global vs. local dynamics So it s common in physics to have several different, but equivalent, views of the same physical situafon Physics of an ideal gas Quantum mechanics, with wave and matrix formulafons We have the same situafon in beam dynamics Global view where we assume a ring or beam line exists and study the global properfes of the system. For example, stability or tune Local view where we worry about the details of the machine What frame of reference is best? What do the fields of our magnets look like in this frame? How do I sftch together the the frames of many magnets? How does a single parfcle move in this system?

22 Equivalent ways of doing dynamics So we have different ways of looking at a beam There are also several ways of doing parfcle dynamics These ways are equivalent to each other and all can be used to solve dynamical problems. The three formulafons of dynamics are 1) Newtonian dynamics 2) Lagrangian dynamics 3) Hamiltonian dynamics The one you should choose depends on the kind of problem you are solving In accelerator physics we tend to use Newtonian and Hamiltonian dynamics, and each one has its own merits. Let s look at each of them in turn

23 Newton Newton s formulafon involves relafng a force to the resulfng mofon using Newton s 2 nd law d dt m ~x = F (~x, ~x, t) Doing the physics means working very hard to figure out the force F, which depends on the parfcle posifons, velocifes and on the independent variable Fme (t) We then solve Newton s second law to figure out the evolufon of the posifon and velocity of a parfcle as a funcfon of the independent variable t

24 A simple harmonic oscillator Think of a mass on a spring Mass m, spring constant k Here, the force on the mass from the spring pushes the mass back to the equilibrium point x F = kx

25 A simple harmonic oscillator The force is given by F = kx When we use Newton s law we get the equafon of mofon m d dtẋ = kx d 2 x dt 2 = k m x This standard equafon has oscillatory solufon (with two constant) x(t) =x 0 sin(!t + 0 )! 2 = k m Which we can write as x(t) =c 1 sin(!t)+c 2 cos(!t) Note we can write the equafon of mofon as m d dtẋ = m!2 x

26 A simple harmonic oscillator with the opposite force The force is given by F =+kx Which leads to the standard equafon of mofon d 2 x dt 2 =+k m x Note the force is now pushing the parfcle to larger amplitude, and we can write the solufon as (again with two constants)! 2 = k m x(t) =c 1 cosh(!t)+c 2 sinh(!t)

27 The Lorentz force law A very important equafon for accelerator physics is the Lorentz force law. This relates the force on a parfcle of charge q from an electric and magnefc field ~F = q( ~ E + ~ẋ ~ B) The vector nature of this equafon is very important, and was covered in courses on electromagnefsm. E and B themselves obey equafons, which specify their vector fields in the presence of sources and boundary condifons. Usually the speed is close to the speed of light and so magnefc fields are efficient to guide parfcles Generally v ~ c in a high energy storage ring. But not always!

28 Hamiltonian mechanics Hamiltonian mechanics is a formulafon of dynamics by Hamilton. This way of working carries the advantage of a well defined mathemafcal framework, where keeping control of variables, invariants and approximafons is much easier than when using Newtonian dynamics. Of central importance is the Hamiltonian funcfon H, which is a funcfon of a set of conjugate variable pairs. For example, in one dimension we would deal with the posifon and the canonical momenta of that parfcle as a conjugate pair

29 Hamilton s equa<ons Given a Hamiltonian which is a funcfon of the canonical variables, the dynamics are worked out using Hamilton s equafons dx i i dp i i This gives a set of 2n first order ODEs to solve, where n is the number of dimensions of the system Conceptually the Hamiltonian plays the role that the force does in Newtonian dynamics. ConvenFonally, students of accelerator physics learn the way with Newton first and the way of Hamilton second. In this course we ll largely adopt this convenfonal approach, of using forces first.

30 A simple example a harmonic oscillator The Hamiltonian for an harmonic oscillator is given by H = p2 x 2m m!2 x 2 In terms of the parfcle posifon and canonical momenta. Hamilton s equafons directly give dx dt = p x m dp x dt = m!2 x Which are equivalent to the equafons of mofon obtained from Newton s formulafon of dynamics. Note the Hamiltonian can be wriqen as the sum of the kinefc and potenfal energies H = T + V

31 The Hamiltonian is a conserved quan<ty If we take the total Fme derivafve of the Hamiltonian which depends on x and p x dh dx dt dp x We can use Hamilton s equafons to rewrite some of the derivafves dh @H @p + If the Hamiltonian does not depend explicitly on Fme, then dh =0 And so H is a conserved quanfty of the system. This is very powerful

32 Hill s equa<on So we ve seen that simple harmonic oscillator equafon, with different signs for the spring constant, can lead to oscillafng and diverging solufons. The basic equafons of mofon for a parfcle in an accelerator are called Hill s equafons, which we will soon derive to be x 00 + Take the second one, which is the equafon of mofon for a region with no bending. Depending on the sign of the constant this corresponds to either sine/cosine or sinh/cosh solufons for z(s). So it seems that we can think of an accelerator, at least linearly, as being made up of piece- wise regions with different sizes and signs of the spring constant. x(s) 00 + k x =0 z 00 kz =0 k(s)+ 1 (s) 2 x(s) =0

33 A mass on a spring with variable k == op<cal system SHM through a long quadrupole d 2 u + Ku(x) =0 ds2 Physical analogies x Pendulum (small angles only!)

34 The basic idea Basically we shall use magnefc and electric fields to accelerate and control the parfcles inside our accelerator ~F = q( ~ E + ~ẋ ~ B) We use magnefc fields to deflect parfcles as we really benefit from The presence of the velocity in the Lorentz force law. Consider a field B =1T The magnefc force on a parfcle of velocity c and charge q can be wriqen as F B = q [m/s] 1[V s/m 2 ] Which we can write in terms of an equivalent electric field (which is high!) F B = q 300[MV/m] Given that making electric fields about 1 MV/m is hard, we really benefit by using magnefc fields to deflect our parfcles. We will, however, use electric fields to accelerate our parfcles (why?)

35 A circular orbit and the design orbit The ideal circular orbit We want to make our parfcles move in a circle, so we apply a constant magnefc field. The force on the parfcle is F B = qvb We equate this to the centripetal force condition for circular orbit: F = mv2 Lorentz force qb = centrifugal force mv F e* v* B F L Zentr mv 0 2 ẑ ρ θ s circular coordinate system p e B = p q B* 2 mv 0 e* v* B

36 The magne<c guide field B = p q! 1 = qb p B =8.3T p = 7000 GeV/c 1 = q 8.3[Vs/m 2 ] [ev/c] = 8.3s [m/s] [m 2 ] 1 = [/m] = [/m]

37 Beam rigidity For the case (normally encountered) of E 0 >> m c 2 We can calculate the beam rigidity from the easy- to- remember equafon B [Tm] = E 0[eV] c[m/s] This can also be wriqen as B [Tm] = 3.3 p 0 [GeV/c] 1 = B[T] p 0 [GeV/c] Which is useful when working with high energy colliders, when the beam energy is normally expressed in units of GeV We will generally normalise magnefc field strengths to the beam momenta, and obtain energy- independent k values k 0 = q p 0 B

38 A quadrupole A linearly increasing field (from the centre of the magnet) ~B = g(z~x + x~z) g Normalised field gradient (this is the number used in nearly all codes) Doesn t depend on energy

39 Exercise parameters of storage ring Imagine a proton storage ring, with a beam momentum of 20 TeV/c and proton bunches, with 1.E10 protons per bunch. 1) What is the stored energy of this machine per beam? 2) If the circumference is 83km, and the field is 6.6 T, what fracfon of ring is filled with dipoles? 3) We would like to build a high energy LHC, with a beam momentum 16 TeV/c. What is the big challenge here? 4) The LHC beam energy is 360 MJ. What problems might this cause?

40 The expansion of the fields We will now figure out the equafons of mofon for a parfcle in an accelerator 1) We will figure out the mofon of the parfcles w.r.t. a design orbit 2) We will assume the deviafon of parfcle coordinates from this design orbit is small, so x,z << the bending radius (2) means that we only need to take into account linear terms in the deviafon of the field B w.r.t. x and z So it makes sense to take some arbitrary guide field and make a Taylor expansion around the design orbit B z (x) =B z0 + db z dx x e p B z(x) = e p B z0 + e p db z dx x + e p d 2 B z dx 2 x ! 1 2 d 2 B z dx 2 x2 + e p e p B z(x) = 1 + kx mx ! ox d 3 B z dx 3 x ! d 3 B z dx 3 x3 +...

41 Oscilla<ons and focusing Consider a constant energy parfcle in a fixed dipole field (into the page) Reference parfcle executes cyclotron mofon Other parfcles of the same energy execute cyclotron mofon of same radius Off- axis parfcle Reference parfcle Compared to reference trajectory, this looks like an oscilla<on. We ll call this a betatron oscilla<on. geometric focusing Number of oscillafons per turn (the tune ) is 1. y x dp/p s y

42 PART TWO- HILL S EQUATION

43 The equa<ons of mo<on in an accelerator We will now derive the transverse equafons of mofon for a parfcle moving under the influence of dipole and quadrupole fields in an accelerator. These are called Hill s equafons. The steps we will follow are 1) Define our curved coordinate system and our coordinates 2) Work out what posifon, velocity and accelerafon vectors look like in this coordinate system. Note we are concerned with the transverse mofon and not concerned with longitudinal mofon at this stage. 3) Use the Lorentz force law to figure out the forces on the parfcle and write down an expression for the change in parfcle momentum 4) Change the independent variable from Fme to space 5) Expand out the equafons so they are linear in x and z (this means we can solve them) 6) Specialise to the case of pure dipole and quadrupole fields The result will be a second order ordinary differenfal equafon in each plane, for the mofon of the parfcle in the machine. Hill s equafons.

44 A storage ring, with a design orbit and dipoles to define it

45 Our coordinate system and our coordinates (x, z, s) The ideal circular orbit We will now develop the equafons of mofon in a linear or circular machine. ẑ We ll use a curved coordinate system, with this curvature produced by a local dipole field. The local curvature is rho The path length along this design trajectory will be s. condition for circular orbit: The posifon vector of a parfcle in this coordinate system is ~R = r~x + y~z where r = + x Lorentz force centrifugal force circular coordinate system The curved reference trajectory is normally called the orbit, and the coordinate system moves with a reference parfcle around the F edesign * v* B orbit defined by the dipoles. F L So our coordinates represent deviafons with respect 2 mv to the design (ideal) orbit, and we 0 Zentr assume these deviafons will be small (x is normally around millimetres). 2 mv 0 For coordinates relafve to this design orbit we use e the * v posifon * B and slope dx/ds=x θ ρ s p e B*

46 Coordinates are with respect to the design orbit The design orbit has curvature rho The path length along this orbit will ulfmately be labeled by s Although we start off with Fme (t) as the independent variable We will generally ignore the longitudinal component of mofon At any point we have a right- handed coordinate system (x, y, s)

47 Kinema<cs in this coordinate system Let s do some kinemafcs in this coordinate system. The velocity is given by So ~v = ~ Ṙ Changes around ring ~v =ṙ~x + r ~x +ż~z ~v =ṙ~x + r ~s +ż~z v x =ṙ v y =ż v s = r tangenfal velocity We can differenfate again for the accelerafon vector: ~R =ẍ~x +ṙ ~x +ṙ ~s + r ~s + r ~s + z~z ~R =ẍ~x +2ṙ ~s + r ~s r 2 ~x + z~z Note we used ~x = ~s ~s = ~x

48 The Lorentz force law To figure out the force on our parfcle, we use the Lorentz force law with no electric field and only transverse magnefc fields, ~x ~y ~s d~p dt = e v x v y v s B x B y 0 r d~p dt = e~v ~ B Expanding the cross product gives the components of the rate of change of p d~p dt = d dt m ~ R = m ~R r 2 = = ev s B y ~x + ev s B x ~y + e(v x B y v y B x )~s EquaFng this to the momentum in terms of our posifon vector R We get an expression for the rate of change of p component by component d~p dt = m ( r~x +2ṙ ~s + r ~s The radial and verfcal equafons of mofon then read eb y m v s z = eb x m v s (assuming gamma is constant, as we only have magnefc fields) r 2 ~x + z~z) Can you prove this? We ve dropped the s parts

49 Changing the independent variable from t to s We now want to change the independent variable from Fme to s, as this is what we care about and magnets are localised in posifon and not Fme. It s wrong to use ṙ = dr dt = ds dt = v s We actually need to include an extra factor due to geometry of the orbit (See Lee) ds dt = v s r = d2 r dt 2 = + x We can now convert the the radial and verfcal derivafves to get v s dr + x ds = v s + x 2 vs d 2 x + x ds 2 z = vs + x 2 d 2 z ds 2 dx ds (need to scale back from orbit to reference orbit, as it s the laqer we want for ds/dt)

50 The equa<ons of mo<on We can use the equafons to change the independent variable from t to s in the equafons of mofon we derived from the Lorentz force law. We do this by swapping each dt for a ds, and then use the fact that r = v s S = Doing this we then obtain the equafon for x (and similarly for z) + x 2 d 2 x ds x = eb y m v s x x 1 2x +... x We then expand the funcfons of x and z, keeping only first order. APer some algebra we obtain the equafons of mofon in x and z d 2 x ds 2 + x 2 = e(b z B z0 ) p d 2 z ds 2 = eb x p 1 = B e p Note we assumed p = m v m v s And we have normalised the field to p/q

51 Equa<ons of mo<on in dipole and quadrupole fields An important applicafon of our transverse equafons of mofon is when the magnefc field contains only dipole and quadrupole components, so we write ~B = B z0 ~z + g(z~x + x~z) g z /@x If we do this, the guide field term B z0 cancels, and we are lep with the equafons of mofon in dipole and quadrupole fields in both planes d 2 x ds 2 + g B x =0 d 2 z ds 2 g B z =0 Note that the fields B z0 and g are periodic funcfons of the independent variable s. This periodicity is either the machine circumference or the length of a repeafng period L. So we have derived our equafons of mofon for transverse mofon. Note some authors play with the signs of the pieces in the second term of the equafon for x. For example, Wille writes k. Be careful!

52 Exercises 1. Prove the magnefc field bends the trajectory of the parfcle but does not do any work on the parfcle. This means the parfcle energy does not change. Hint : d(gamma)/dt=0 2. Why are magnefc forces necessarily transverse? What does this mean about longitudinal control of a beam?

53 So we have Hill s equa<ons So, aper lots of effort, we derived Hill s equafons. They are linearised second order differenfal equafons for the transverse variables x and z in dipole and quadrupoles fields in a parfcle accelerator. d 2 x ds 2 + x(s) 00 + g B x =0 k(s)+ 1 (s) 2 d 2 u + Ku(x) =0 ds2 K>0 d 2 z ds 2 x(s) =0 g B z =0 In the equafon for x, g comes from the quadrupoles and there is also natural focusing from the dipoles in the plane of curvature. We assumed that k and rho are constant in the solufon, so we assume these funcfons of s are piecewise constant K<0 for u = z for u = x = const k = const We can compactly write Hill s equafons in both planes, denofng x or z with u and wrifng the posifon dependent spring constant as K

54 Effec<ve magnet lengths B *l In reality, magnets tend to have fields which extend beyond the physical length of the magnet. We account for this using an effecfve length, which is the length of the magnefc elements as seen by the beam. B lef f = Z lmag Bds 0

55 Solu<on of piece- wise Hill s equa<ons x 00 + k x =0 These are our equafons of mofon, the horizontal plane and the verfcal plane z 00 kz =0 x 00 + K x =0 K = k K = k These equafons look like the equafon for a simple harmonic oscillator, if we write K for the constant and assume it s constant over the region of the accelerator we are interested in So we know the solufon from out studies of a mass on a spring! Let s guess at K>0 x(s) =c 1 cos( p Ks)+c 2 sin( p Ks) Take the derivafves and subsftute into the equafon of mofon x 0 (s) = c 1! sin( p Ks)+c 2! cos( p Ks) x 00 (s) = c 1! 2 cos( p Ks) c 2! 2 sin( p Ks)=! 2 x(s) p K =!

56 Solu<on of piece- wise Hill s equa<ons So this solufon seems to work. Of course we assumed K > 0 to get an oscillatory solufon but let s worry about the case of K < 0 later. What about those two unknown constants? We can fix these from the inifal condifons x(0) = x 0! c 1 = x 0 x 0 (0) = x 0 0! c 2 = x0 0 p K So we can write down the evolufon of the variables x and x in a region with constant and posifve K as K>0 x(s) =x 0 cos( p Ks)+x p K sin( p Ks) x 0 (s) = x 0 p K sin( p Ks)+x 0 0 cos( p Ks)

57 uadrupole K > 0: 1 xs () x0 cos( Ks) x0 sin( Ks) Solu<on of piece- wise K Hill s equa<ons What does this mean? NoFce that the x variable receives a negafve kick, which corresponds to poinfng the parfcle more towards the axis x( s) x K sin( K s) x cos( K s) 0 0 ce expressed in matrix formalism: s = s 0 s = s 1 M foc * x x s0 Also note that the final coordinates are linear combinafons of the inifal coordinate, which is a direct consequence of linearising the equafons of mofon. 1 cos( Ks) sin( K K sin( K s) cos( K s) Ks This means we can write the solufon very compactly as a matrix equafon M foc quad = 0 x x 0 1 = M quad x x 0 cos( p 1 Ks) p K sin( p! Ks) p p p K sin( Ks) cos( Ks) 0

58 A defocusing quadrupole What about the case of K < 0? This has equafon of mofon x 00 K x =0 And corresponds to a diverging solufon. We remember our studies of a mass on a spring and use the sinh/cosh funcfons f(x) = cosh(x) f 0 (x) =sinh(x) Which means the general solufon to the equafon of mofon is M defoc quad = x(s) =c 1 cosh( p K s)+c 2 sinh( p K s) hor. defocusing quadrupole: This corresponds to a defocusing lens, and we x K *x 0 can write the matrix in the same kind of form but using sinh/cosh funcfons Remember from school: s = 0 cosh( p 1 K s) p sinh( p K s) K f ( s) cosh( s), f ( s) sinh( s) + p K sinh( p K s) cosh( p K s) Ansatz: x s) a cosh( s) a sinh( ) ( 1 2 s! s = s1

59 Solu<on of piece- wise Hill s equa<ons x 00 + k x =0 z 00 kz =0 These are our equafons of mofon, the horizontal plane and the verfcal plane This means for a given quadrupole magnet with no curvature, we have one sign of the constant K in one plane (say x) and the opposite sign in the other plane. So a quadrupole, really because of Maxwell s equafons and the need that the magnefc field is curl free, focuses in one plane and defocuses in another. So by convenfon we call a horizontally focusing quadrupole a focusing quadrupole and vice versa. e p db z dx = g B = k Therefore a quadrupole with a posifve sign for db z /dx gives k > 0, and so is horizontally focusing. Therefore dbz/dx < 0 is horizontally defocusing.

60 The dri: space x x L A drip space is a region of the beam line with no electromagnefc fields. We can figure out the evolufon equafons for x and x by either simple geometry or taking a limit of the quadrupole matrices for K- >0. Either way we find the variables change as x(s) =x 0 + x 0 0 L x 0 (s) =x 0 0 Which we can write as a matrix very easily M drift = 1 L 0 1

61 The thin lens approxima<on We know the matrix of a quadrupole can be wriqen as M foc quad = cos( p 1 Ks) p K sin( p! Ks) p p p K sin( Ks) cos( Ks) Very open the lens is short compared to the focal length This means we can take the limit of very short magnet, whilst keeping the focal length constant: This give us the thin lens matrices, which are very useful for quick calculafons of a given accelerator structure (recall k < 0 for a horizontally focusing quadrupole) M thin = f 1 1 f = Kl q

62 Now we know four linear transport matrices M foc quad = cos( p 1 Ks) p K sin( p! Ks) p p p K sin( Ks) cos( Ks) M defoc quad = cosh( p 1 K s) p sinh( p K s) K + p K sinh( p K s) cosh( p K s)! M drift = M thin = 1 L f 1 How would you turn these into a simple tracking code? When would your code give valid results?

63 But we have lots of elements? In a real accelerator we have lots of elements in a line. M cell = M QF M bend M QD M drift M QF x x 0 s=1 = M(s =1,s= 0) Each element is represented by a matrix. How do we transform through these sequences of elements? Easy! We mulfply the matrices of each element to give an overall transfer matrix through the system x x 0 s=0 First element! Note, to compute the parfcle we need to know the fields and magnet posifons very well!

64 The dipole bend assump<on of a curved coordinate system Is there a transfer matrix for a dipole? Yes, there is. First of all, we used a curved coordinate system so the design trajectory follows The curvature of the magnet

65 A pure dipole To get the matrix, we start from the matrix of a quadrupole M foc quad = cos( p 1 Ks) p K sin( p! Ks) p p p K sin( Ks) cos( Ks) In the horizontal plane, we have a pure bending contribufon to K And so we get for a pure bend K = 1 2 = l/ We can see that there is geometric focusing in the plane of the bend The matrix in the verfcal plane is a drip What s the matrix for an angle of 2Pi? Is it sensible?

66 The doublet Consider a double, which the the simplest example of a system with two quadrupoles Focal lengths f 1 f 2 SeparaFon d The thin lens map for a single quadrupole is now known, as is the drip map If we mulfply these matrices, the focal length for the combined system is easily obtained from (2,1) element of the composite matrix 1 f = 1 f f 2 d f 1 f 2 If I let f 1 = - f 2 then the leading terms cancel and the doublet is focusing in both planes at the same <me. This is a very pleasant feature! This feature led to the invenfon of the alternafng gradient (or strong focusing) principle and is the basic building block of modern accelerator lajces. Can you see in terms of tracing rays through the two lenses why it happens?

67 Maps We have quietly brought in the idea of a map The matrix M is that map that brings an inifal state vector to a final state vector X(s 1 )=M(s 1 s 0 )X(s 0 ) For the linear case, the map can be represented as a matrix, so we have find a matrix representafon of the map Matrix representafon ß à Linear system For non- linear systems, matrices do not work any more and we need to find new representafons of the maps, for example Taylor maps or Lie maps. This, however, is a liqle beyond this course.

68 Matrix algebra and transfer matrices We know how to combine matrices, with a standard rule M(s 2 s 0 )=M(s 2 s 1 )M(s 1 s 0 ) Matrix algebra is not commutafve But it is associafve and we can form matrix sub- groups (Provided we maintain the order of the matrices!!) One parfcularly useful map is the one- turn map If we start at locafon s in a ring of circumference C, then the one turn map is defined as one turn around the ring M(s + C s) This means the map for N revolufons of the ring is found from N applicafons to a given parfcle state vector of the one turn map More on this idea later M(s + C s) N

69 Why are matrices useful? We ve basically re- wriqen our equafons in terms of matrices. Is this useful? Yes, as it means we can use all the formal machinery of linear algebra e.g. Matrix mulfplicafon eigenvalues eigenvectors Traces Similarity transforms Plus quickly see the impact on the beam of a series of elements. All this is very powerful

70 So how do we do it in a code? Accelerator codes simply assume a piecewise- confnuous representafon of the accelerator structure. but because of edge focusing the number of matrices is not the same as the number of elements.

71 Real lafces and lafce design Remember that when we design lajces, that eventually it will get built (hopefully!) Reality uses up more space than ideal elements.

72 A single par<cle for a single turn And, now, we have a parfcle in our accelerator for a few elements

73 Question: what will happen, if the particle performs a second turn? What if it wants to make many many turns?... or a third one or turns x turn for beam 1 0 IP5 IP6 IP7 IP8 IP1 Injected beam + beam after one turn on screen So lots of parfcles over many turns form a kind of envelope of the beam. s l position [ km ] Horizontal trajectory Note we can see the straight reference parfcle in this plot, which is not a real parfcle.

74 PART THREE- THE LATTICE FUNCTIONS

75 Back to Hill s equa<on x(s) 00 + k(s)+ 1 (s) 2 Hill s equafon is a second order differenfal equafon for a system with periodic focusing properfes It s like a pendulum but the restoring force is not a constant (like gravity is a constant on a pendulum) In fact, the variable spring constant k(s) depends on the magnefc properfes of the ring If this ring has periodicity L, then so does the funcfon k(s) k(s + L) =k(s) x(s) =0 So we can expect a kind of quasi- harmonic oscillafon, where the frequency and amplitude depend on the locafon in the ring and show periodicity similar to that of the funcfon k(s)

76

77 The Courant- Snyder formalism In the Courant- Snyder formalism we assume a solufon of Hill s equafon inspired by our intuifon on posifon dependent amplitude and phase. It is this. x(s) = p (s) cos( (s)+ 0 ) β(s) has the physical meaning of an amplitude which depends on the posifon around the accelerator. Ψ(s) is a posifon dependent phase ε is a constant. Because Hill s equafon is linear, the constant does not appear in it. We ll see later that ε is special and is called the emiqance β(s) is the key quanfty in the Courant- Synder formalism and has many names : the beta funcfon, the beam envelope funcfon, the Courant- Synder beta funcfon, the amplitude funcfon and so on. It is always posifve. We ll see that it represents the focusing properfes of a lajce, and a small beta funcfon means a Fghtly focused lajce, and vice versa. The periodicity of the magnefc system is very important, and we will see (s + L) = (s)

78 A differen<al equa<on for the beta func<on If we take the derivafves of the Courant- Synder ansatz and subsftute into the equafon of mofon, we find we get two terms, one proporfonal to cosine and one proporfonal to sine. This is a good exercise to do! The coefficients of these terms must vanish separately, and we eventually obtain two differenfal equafons 1 2 ( ) = =( 0 ) 0 0 = k =0 The second equafon equafon can be integrated immediately, since And we are free to choose the integrafon constant to be unity

79 The Courant- Synder (or lafce) func<ons We then immediately have the result for the phase funcfon Z s (s) = d (s) ds (s) = ds (s) So this posifon dependent phase is related to an integrafon of the beta funcfon along the beam line, and knowing the beta funcfon means we can compute the phase funcfon We can now eliminate the phase funcfon from the first of the differenfal equafons to get a differenfal equafon for the beta funcfon k =1 So knowing the distribufon of focusing strengths along a beam line determines beta, although we never actually solve this equafon in pracfce! Finally, we define the two funcfons (with beta, called the lajce funcfons) (s) = 1+ 2 (s) (s)

80 The appearance of the ellipse Once the beta funcfon is known, and hence alpha and gamma, the mofon of a single parfcle is completely specified by specifying the emiqance and the inifal phase factor of the parfcle. So we have x 0 (s) = x(s) = p (s) cos( (s)+ 0 ) p p (s) cos( (s)+ 0 )+sin( (s)+ 0 ) We can combine these two equafons to give the quanfty x 0 + x = p sin( + 0 ) Which means we can write an expression which is invariant for a parfcle x 2 +( x 0 + x) 2 = Or, expanding the square, we arrive at the famous result x 2 +2 xx 0 + x 02 =

81 The emi`ance is an invariant Let s look at this equafon carefully. x 2 +2 xx 0 + x 02 = For every point in the accelerator we have a value of the funcfons alpha, beta and gamma. They depend on the lajce. At at parfcular point, if we combine the parfcle posifon and angle with these lajce funcfons we get an invariant, which was the emiqance we saw in the solufon to Hill s equafons in the Courant- Synder formalism As the parfcle moves to the next locafon in the accelerator, where we have different lajce funcfons, it get a different posifon and angle. However if we form this funcfon again at the new locafon we get the same value as before. In other words, the emiqance is a constant of the parfcle. So the area of the ellipse is the area in (x,x ) space covered by the parfcle. This is constant. This is a very important concept.

82 The ellipse x 2 +2 xx 0 + x 02 = This funcfon actually describes an ellipse in the (x,x ) plane, with ellipse parameters described by the values of alpha, beta and gamma. Beta controls the extent along the x axis, gamma controls the extent along the x axis, alpha tells you how upright the ellipse is (example what values of alpha, beta and gamma give you a perfect circle?) Exercise: If a parfcle sits on an ellipse, which way around the ellipse does it move? The area of the ellipse is given by A = Which means the area of an ellipse transcribed by a given parfcle is constant

83 A stroboscopic plot of a par<cle turn a:er turn a:er turn Recall that the lajce parameters are funcfons of the focusing of the lajce, so every point in the lajce has a value of the lajce funcfons and so every point in the lajce has it s own orientafon of the ellipse. A given parfcle has it s own value of the emiqance, so sejng the area of the ellipse it moves around. Let s play a game we sit at a fixed point in the ring and watch a single parfcle turn aper turn aper turn. This can be done with a simple computer code. (y 1, y 1 ) (y 2, y 2 ) (y 3, y 3 ) (y 4, y 4 ) All these points lie on the ellipse. Note the parfcle jumps around the ellipse and does not move around it confnuously

84 Moving along a beam line s 3 x s 1 x s 2 x x x x ( β 3, α 3, ϒ 3 ) ( β 2, α 2, ϒ 2 ) ( β 1, α 1, ϒ 1 )

85 The beta func<on The beta funcfon is a central quanfty in the Courant- Synder formalism It is a posifve funcfon of posifon in the machine, and carries the same periodicity that the lajce itself carries. It is determined by the focusing properfes of the lajce, and is a funcfon which is roufnely computed in the design and operafon of parfcle accelerators. It is maximised in a focusing quadrupole and minimized in a defocusing quadrupole. Let us now look at some examples.

86 The beta func<on of the LHC Beta functions for IP1 and IP5 S. Redaelli, LPCC lectures, 07/ β = 55 cm σ = 16 μm; D = 0m 89

87 Optics: correction of β functions And you can measure the beta func<on too! R. Tomàs

88 Determina<on of the transfer matrix in terms of the Courant- Snyder func<ons Let us now write a general transfer matrix between two points in terms of the lajce funcfons at these two points. To begin with, we return to the Courant- Synder form of the solufon to Hill s equafon, note it depends on two constants and write this ansatz in a slightly different form Where c 1 and c 2 are constants yet to be determined. If we define the inifal condifons at the point 0 to be (0) = 0 (0) = 0 (0) = 0 And write the inifal parfcle coordinates to be x 0 and x 0 then we can fix the unknown constants to be c 1 = x 0 p 0 And so we can write x(s) in the form x(s) = s (s) 0 c 2 = p 0x p 0 [cos (s)+ 0 sin (s)]] x 0 + p 0 x 0 (s)x 0 0 sin (s)

89 Determina<on of the transfer matrix in terms of the Courant- Snyder func<ons We see the expression for x(s) is linear in x 0 and x 0. x(s) = s (s) 0 [cos (s)+ 0 sin (s)]] x 0 + p 0 (s)x 0 0 sin (s) Taking the derivafve of this expression, we can cast this equafon into a convenient matrix form (as it s linear) (I m not going to write down the derivafve consider it an exercise) where M(s 1 s 0 )= x(s1 ) x 0 (s 1 ) q 0 1 p 1 0 = M(s 1 s 0 ) 1 (cos + 0 sin ) 0 cos p 1 0 sin x(s0 ) x 0 (s 0 ) This means the transfer matrix between two points is purely determined by the lajce funcfons at each point and the phase advance between the points!! q p 1 0 sin 1 (cos 0 sin ) 0 1 A = (s 1 ) (s 0 )

90 The one- turn map The one turn (strictly one period) map is a very important quanfty. StarFng with M(s 1 s 0 )= q 0 1 p (cos + 0 sin ) 0 cos p 1 0 sin q p 1 0 sin 1 (cos 0 sin ) 0 1 A The map for one turn of the ring means we come back to the same posifon, and so 1 = 0 = 1 = 0 = 1 = 0 = And so the one turn maps is M(s + L s) = cos + sin sin sin cos sin The phase advance for the one turn is given by 1 0 = (s) = 1+ 2 (s) (s)

91 The one- turn map We can use this map to figure out the lajce funcfons. If we mulfply all the matrices for all the elements in the ring together, we obtain the total matrix for one turn of the machine (again, strictly, one period) = m 12 sin M = m11 m 12 m 21 m 22 We can get the one turn phase from the trace of this matrix, comparing it to the form we have for the one turn map m11 + m 22 = arccos 2 We can get the lajce funcfon from the other matrix elements. = m 11 m 22 2sin TrM apple2 = m 21 sin Note for the phase advance to be real valued and hence stable, we need

92 An exercise Using the expressions for the lajce funcfons in terms of the matrix elements, show that the condifon det(m)=1 is the same as (s) = 1+ 2 (s) (s) by using the expression M(s 1 s 0 )= q 0 1 p (cos + 0 sin ) 0 cos p 1 0 sin q p 1 0 sin 1 (cos 0 sin ) 0 1 A (Understand this by seeing that a 2x2 matrix has 4 free parameters if real. If we require det M=1, then this drops to 3. If we parameterise the matrix with the three lajce funcfons and the phase advance, then the equafon for gamma drops these four parameters to three.)

93 The one- turn map at a different loca<on Imagine we know the one turn map at one locafon, say s. Is there a way to figure it out at another locafon, say s, provided we know the transfer matrix M for s to s? The answer is yes. They are related to each other by a similarity transform, and M(s 0 + C s 0 )=M(s 0 s) M(s + C s) M 1 (s 0 s) I am going to state this without proof (which I avoid as much as possible!). Similarity transforms come from matrix theory and all manner of nice properfes such as idenfcal eigenvalues and traces before and aper the transformafon. (We don t use them very much in this introductory course but you will later) Let s be concrete and denote the matrix M (from s to s ) by M = m11 m 12 m 21 m 22 Let s use this to figure out how the lajce funcfons transform from place to place if we know the transfer matrix.

94 The transforma<on of the lafce func<ons StarFng with the similarity transform, M(s 0 + C s 0 )=M(s 0 s) M(s + C s) M 1 (s 0 s) We can express the one turn maps in terms of the lajce funcfons at the locafons s and s M(s + L s) = cos + sin sin sin cos sin And do about 1 page of algebra to obtain the lajce funcfons at point s (or 1) in terms of the lajce funcfons at point s (or 0) and the elements of the matrix M. The answer is A = m 11 m 22 + m 12 m 21 m 11 m 21 m 12 m 22 2m 11 m 12 2m 21 m 22 m 2 11 m 2 21 m 2 12 m 2 22 So knowing M, we can transform the lajce funcfons to any point in the beam line. Needless to say, this expression is important!!! A

95 The phase advance and tune Several Fmes we have used the phase advance for one turn of a ring (period). It is Z s+l = s = 2 = ds (s) And so given by an integral over the beta funcfon. We open call the phase advance for one turn of a ring the tune, and express it in units of 2*pi Z s+c (or Q) s ds (s) There is one tune for each plane, including the longitudinal plane, and it s a very important funcfon for beam stability. Note we can evaluate the tune at any point in the ring and always get the same answer (a property not shared by alpha, beta and gamma) (This is because of the trace is invariant under similarity transformafons)

96 Tune = 2 = Z s+c s ds (s) 1 turn"

97 Tune Tune is the number of oscillafons made per turn What is measurable and relevant is the non- integer part of the tune.28.31

98 Courant- Snyder parameter evolu<on in a dri: In a drip space of length L we have M drift = 1 L 0 1 And so m 11 =1 m 12 = L m 21 =0 m 22 =1 The lajce funcfons evolve 1 = L + 0 L 2 1 = 0 0 L A parfcle evolves x(l) =x 0 + Lx 0 0 x 0 (L) =x = 0 Is a drip stable?

99 Exercise Consider a thin lens quadrupole with focal length f. Work out the change in the lajce funcfons through this quadrupole. Contrast the behaviour of a focusing (f > 0) and a defocusing (f < 0) quadrupole on the change in the beta funcfon through the lens. (Hint: look at how alpha changes). As a result, where would you expect to find the maximum horizontal beta funcfon in a beam transport channel composed of alternafng focusing and defocusing quadrupoles?